The Mole Concept and Stoichiometry

2MnO₄⁻(aq) + 5H₂O₂(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5O₂(g) + 8H₂O(l)[1]

Volume of FA 2 = 8210 × 25.0 = 205 cm³[1]

205.00 cm³ of FA 2 would be needed from the burette, so the titration would be very tedious, as the burette would have to be refilled carefully several times.[1]

Amount of H₂O₂ in 25.0 cm³ of FA 3 = 52 × y1000 × 0.0200 = 5.00 × 10⁻⁵y mol[1]

Amount of H₂O₂ in v cm³ of FA 1 = amount in w cm³ of FA 3 = w25.0 × 5.00 × 10⁻⁵y = 2.00 × 10⁻⁶wy mol

[H₂O₂] in FA 1 = 2.00 × 10⁻⁶wy ÷ v1000 = 0.00200 wyv mol dm⁻³[1]

Amount of O₂ from 1 dm³ = 13.422.7 = 0.5903 mol

Amount of H₂O₂ = 2 × 0.5903 = 1.181 mol, so [H₂O₂] = 1.18 mol dm⁻³[1]

NoteThe volume is at s.t.p., so use 22.7 dm³ mol⁻¹, not 24.0. The mole ratio comes from the decomposition 2H₂O₂ → 2H₂O + O₂, not from the equation in (a).

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O   Oxidation: C₂O₄²⁻ → 2CO₂ + 2e⁻

Overall: 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O[1]

Amount of S₂O₃²⁻ = 24.501000 × 0.20 = 4.90 × 10⁻³ mol

Amount of I₂ = ½ × 4.90 × 10⁻³ = 2.45 × 10⁻³ mol

Amount of remaining MnO₄⁻ in 25.0 cm³ = 25 × 2.45 × 10⁻³ = 9.80 × 10⁻⁴ mol[1]

Remaining MnO₄⁻ in 100 cm³ = 9.80 × 10⁻⁴ × 10025.0 = 0.00392 mol

Initial MnO₄⁻ = 100.01000 × 0.10 = 0.0100 mol, so MnO₄⁻ reacted with C₂O₄²⁻ = 0.0100 − 0.00392 = 0.00608 mol[1]

Amount of CaC₂O₄ = 52 × 0.00608 = 0.0152 mol

Mass of pure CaC₂O₄ = 0.0152 × [40.1 + 2(12.0) + 4(16.0)] = 1.95 g

% purity = 1.953.20 × 100% = 60.9%[1]

No. of moles of EDTA⁴⁻ = 0.0200 × 35.451000 = 7.09 × 10⁻⁴ mol[1]

No. of moles of Al³⁺ in 25.0 cm³ = 7.09 × 10⁻⁴ mol

No. of moles of Al³⁺ in 250 cm³ (the whole water sample) = 7.09 × 10⁻⁴ × 25025 = 7.09 × 10⁻³ mol[1]

HClO + 2I⁻ + H⁺ → Cl⁻ + I₂ + H₂O[1]

Amount of S₂O₃²⁻ = 6.01000 × 0.0050 = 3.0 × 10⁻⁵ mol

Amount of I₂ = amount of HClO = amount of N-chlorosuccinimide = 1.5 × 10⁻⁵ mol[1]

M_r(C₄H₄O₂NCl) = 133.5, so mass = 1.5 × 10⁻⁵ × 133.5 = 2.0 × 10⁻³ g[1]

[R]: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (× 1)   [O]: Fe²⁺ → Fe³⁺ + e⁻ (× 6)

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O[1]

Amount of Cr₂O₇²⁻ reacted = 21.451000 × 0.100 = 2.145 × 10⁻³ mol

Since 6Fe²⁺ ≡ Cr₂O₇²⁻, amount of Fe²⁺ in 25.0 cm³ of solution = 6 × 2.145 × 10⁻³ = 0.01287 mol[1]

Amount of Fe²⁺ in 250 cm³ of solution = 0.01287 × 25025.0 = 0.129 mol[1]

By the law of conservation of mass, amount of Fe in the iron ore = amount of Fe²⁺ in 250 cm³ of solution.

Mass of Fe in the iron ore sample = 0.1287 × 55.8 = 7.181 g[1]

% by mass of Fe in the sample = 7.18111.15 × 100% = 64.4%[1]

Let the mass of a sample of R be 100 g.

Amount of Cs = 58.1 ÷ 132.9 = 0.437 mol

Amount of P = 6.78 ÷ 31.0 = 0.219 mol

Amount of S = 35.1 ÷ 32.1 = 1.09 mol

Simplest ratio Cs : P : S = 2 : 1 : 5, so the empirical formula is Cs₂PS₅[1]

Let the chemical formula be (Cs₂PS₅)_n.

(2(132.9) + 31.0 + 5(32.1))n = 914.6

457.3n = 914.6, so n = 2

The chemical formula of R is Cs₄P₂S₁₀[1]

P₂S₁₀⁴⁻[1]

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)[1]

Amount of Fe₂O₃ used = 150 ÷ 159.6 = 0.940 mol

Amount of CO used = 60 ÷ 24.0 = 2.50 mol

Since 1 mol of Fe₂O₃ reacts with 3 mol of CO, 0.940 mol of Fe₂O₃ requires 2.82 mol of CO for complete reaction. Hence, CO is the limiting reagent.[1]

Since 3 mol of CO form 2 mol of Fe, the maximum amount of Fe formed = 2.50 × 23 = 1.667 mol

Maximum mass of Fe = 1.667 × 55.8 = 93.0 g[1]

Amount of CO₂ formed = amount of CO used = 2.50 mol

Number of CO₂ molecules = 2.50 × 6.02 × 10²³ = 1.51 × 10²⁴[1]

Percentage yield = 7093.0 × 100% = 75.3%[1]